JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A solid sphere of mass ' \(m\) ' and radius ' \(r\) ' is allowed to roll without slipping from the highest point of an inclined plane of length ' \(L\) ' and makes an angle \(30^{\circ}\) with the horizontal. The speed of the particle at the bottom of the plane is \(v_1\). If the angle of inclination is increased to \(45^{\circ}\) while keeping \(L\) constant. Then the new speed of the sphere at the bottom of the plane is \(v_2\). The ratio \(v_1^2: v_2^2\) is:
- A \(1: \sqrt{2}\)
- B \(1: \sqrt{3}\)
- C \(1: 3\)
- D \(1: 2\)
Answer & Solution
Correct Answer
(A) \(1: \sqrt{2}\)
Step-by-step Solution
Detailed explanation
using WET \(\begin{aligned} & \mathrm{W}_{\mathrm{g}}=\mathrm{k}_{\mathrm{f}}-\mathrm{k}_{\mathrm{i}} \\ & \mathrm{Mg} \mathrm{~L} \sin \theta=\mathrm{k}_{\mathrm{f}}-\mathrm{k}_{\mathrm{i}} \end{aligned}\) K.E. in pure rolling…
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